In one of our Brain Potions from last year (found here), we explored a betting strategy to supposedly win "Infinite Money" playing a version of Roulette.

In this week's Brain Potion we present a simple card game and explore how much money we can win!

Start with a deck of $2N$ cards, with $N$ black cards and $N$ red cards and start with $\$1.00$. The dealer of the game flips the cards over, one by one, until all $2N$ cards are dealt. Before he flips over each card, you have a chance to bet/predict the color of the next card. You can choose to risk some/all/none of your money, and if you are correct you win double what you bet.

Example: if you have $\$\dfrac{2}{3}$ and choose to bet $\$\dfrac{1}{6}$, then if you are wrong you end up with $\dfrac{2}{3} - \dfrac{1}{6} = \$\dfrac{1}{2}$ while if you are correct you end up with $\dfrac{2}{3} - \dfrac{1}{6} + \dfrac{2}{6} = \$\dfrac{5}{6}$.

Your goal is to come up with a strategy to guarantee you win as much money as possible, leaving nothing to chance.

Example: if $N=1$ we have one red $R$ and one black $B$ card. We can guarantee we finish with $\$2$ using the following strategy. Bet nothing before the first card is flipped. In this way you learn the result of the first card and can bet your full $\$1$ on the second card (and are guaranteed to be correct!). This gives you $1 + 1 = \$2$ at the end.

Try to come up with a strategy for $N = 2$. That means you have to have a strategy that deals will all possible ways the cards could be dealt:
$$BBRR, BRBR, BRRB, RBBR, RBRB, RRBB$$
so that you always end up with the same amount of money. Hint: You can guarantee you win more than $\$2$!

As a challenge, once you've solved the problem for $N=2$, try $N=3$, etc. There is a nice pattern for how much money you can win, depending on $N$. Try to find (and prove) it!

Please share any thoughts or questions you have below. We'll monitor the responses and give our thoughts as well! Have your own request, idea, or feedback for the Brain Potion series? Share with us in our Request and Idea Thread available here.

In the above question we asked for a strategy so you could guarantee you won as much money as possible. This week we explore this game a little further, introducing the concept of Expected Value and showing how expected values relate to our card game!

Let's consider some simple examples first.

Example: Pretend you usually get $\$10$ a week for your allowance. One day your mom comes to you with a proposal. Instead of your current $\$10$ a week, each week you will flip a (fair) coin. If the result is heads, you get $\$20$, while if the result is tail you get $\$0$. Of course this new way of getting your allowance is risky, but is it really better or worse than your old allowance in the long run?

The concept of expected values can explain why the new allowance proposal is very similar in the long run to your previous allowance.

Originally you got $\$10$ each week, therefore each week you expect $\$10$!

Under the new proposal there is a $50\%$ chance you get $\$20$ and a $50\%$ chance you get $\$0$. Thus we say the expected value of your allowance is
$$50\%\times \$20 + 50\%\times \$0 = \$10,$$
the same as before!

Example: Enjoying the chance involved in your mom's proposal, you suggest your own method for determining your allowance! Each week you'll roll a fair six-sided die. If the result is $2$, $3$, or $4$ you'll get $\$10$, if the roll is $5$ or $6$ you'll get $\$20$, but, if the roll is $1$ you'll give your mom $\$10$! What is the expected value in this case?

With this proposal there is a $\dfrac{1}{6}$ chance you lose $\$10$, a $\dfrac{3}{6}$ chance you get $\$10$, and a $\dfrac{2}{6}$ chance you get $\$20$. Hence the expected value is
$$\frac{1}{6}\times (-\$10) + \frac{1}{2}\times \$10 + \frac{1}{3}\times \$20 = \$10,$$
still the same as before!

Let's now use expected values to explore the card game above in more detail when $N = 2$. Recall this means, when shuffled, the $2$ black ($B$) cards and $2$ red cards ($R$) can be arranged in $6$ ways, shown below:
$$BBRR, BRBR, BRRB, RBBR, RBRB, RRBB.$$
Each of these $6$ orderings is equally likely, so has a $\dfrac{1}{6}$ chance of occurring.

You're still trying to figure out how to play the game (as asked last week) to make sure you always win the same amount of money each time you play. However, your two friends Rick and Mark have other plans.

Rick wants to go all or nothing. He always thinks that the cards will come in the order $BBRR$, so will always bet $B$ for the 1st card, $B$ for the 2nd card, etc, risking ALL of his money each time.

Mark plays it a little safer. He only bets on the 2nd and 4th card. He still bets all his money, but bases his guess on earlier cards. For example, if the 1st card is $B$, then he bets all on $R$ for the 2nd. (Note Mark will always get the 4th card correct.)

Rick has the chance to win up to $\$16 = \$1\times 2\times 2\times 2\times 2$ while Mark has the chance to win up to $\$4 = \$1\times 2\times 2$. However, the expected values of what they win are actually equal. What is this expected value?

Further, the amount you can guarantee you win is equal to this expected value as well. Try to use this fact to help you come up with a strategy! See if you can extend the reasoning about expected values to help with $N = 3$ or above in the game as well!

Please share any thoughts or questions you have below. We'll monitor the responses and give our thoughts as well! Have your own request, idea, or feedback for the Brain Potion series? Share with us in our Request and Idea Thread available here.

If there are \( N \) cards of each color then there are \( \binom {2N} {N} \) possible orders for the colors of the cards. If one were to bet 100% on each card it is possible to win \( $2^{2N} \), but only if you guess the exact order, any mistake leaves $0. So one strategy is to assign \( \frac{1}{ \binom {2N} {N} } \) of your original dollar to each of the potential card color orders, then use the 'bet it all' strategy separately and simultaneously with each equal part of the dollar. All except one are guaranteed to lose, but the one that wins will have \( \frac{2^{2N}}{ \binom {2N} {N }} \).

This yields 8/3, 16/5, 128/35 for N=2, 3, 4, and continues to grow unbounded with N.

We're not allowed to make multiple bets in a round, but we can execute the strategy 'virtually.' If, on any round, there are \( p \) red and \(q+p\) black cards (WLOG assume \( q \geq 0 \) ), each simulation that is still alive will have the same amount of money, and we are guaranteed that at least p will win and p will lose. So we can get the same net affect by betting \( \frac{q}{q+2p} \) of our total money on black. For example, in the first round we bet zero, and in round 2 we bet \( \frac{1}{2N-1} \) on the opposite of the first color.